[aquafox]

If I tell you I have function f with f(a) = 10 and on it's path from a to b, the graph first increaes by 5 units then by another 10, and then later on drops by 25 units, you can immediately deduce that f(b) = f(a) + (+5 +10 -25) = 0. The fundamental theorem of calculus uses the same concept:

To see why \int_a^b f(x) dx = F(b) - F(a) with F'(x) = f(x),

we replace f with f' (and hence F with f) and get

\int_a^b f'(x) dx = f(b) - f(a).

Re-arranging terms, we get

f(b) = f(a) + \int_a^b f'(x) dx.

The last line just says: The value of function f at point b is is the value at point a plus the sum of all the infinitely many changes the function goes through on its path from a to b.

[dist-epoch]

how does this work for these two different paths which arrive at the same endpoint:

+5 +10 +0 -25 = -15

+5 +12 -2 -25 = -15

They have different graphs if you consider the values above sampling points, this is what parent is asking.

[nh23423fefe]

the derivative becomes a telescoping sum. it doesn't matter how many points you insert or what the values are because x-x=0

let x_i be in (a,b) with any i drawn from [0,N] and x_0=a and x_N=b

then

int_a^b d/dx F = (F(x_1) - F(x_0))+ (F(x_i+1) - F(x_i))+ (F(x_n) - F(x_n-1)) = F(x_N) - F(x_0)

the derivative produces a ratio of differences and the integral is a weighted sum of those same differences

every middle term is added and subtracted. and only the endpoints remain with opposite sign